The efficiency of water splitting into H2 and O2 is not very high. The efficiency means the ratio of energy in H2 (usually the heat released after burning) to electrical energy input. What is the limiting factor during this electrolysis process? How can we improve the efficiency?
The conversion efficiency means the ratio of energy in H2 (usually the heat released after burning) to electrical energy input.
You need a higher voltage for a practically viable electrolyser (say 1.9-2.2 V for alkaline electrolysis) than you would expect from thermodynamics, which would suggest that you could get away with 1.23 V in standard conditions. However, this reversible potential is a thermodynamic limit and the reaction happens infinitely slowly... if you want a reasonable production rate, you need to push it harder to increase the driving force by applying additional voltage. The extra 0.7 V reduces current efficiency, but enables you to get a decent current density: there's no point having an electrolyser that produces a micro-mol per day after all. Economics and thermodynamics don't always mix.
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